Flat morphism preserves associated points

Let $f:X\to Y$ be a flat scheme morphism, let $\mathcal F\in \operatorname{QCoh}(Y)$.

Let $\operatorname{Ass}_Y(\mathcal F)$ be the set of associated points to $\mathcal F$.

Is it true that $Ass_X(f^{*}\mathcal F) = f^{-1}(Ass_Y(\mathcal F))$?

I couldn't find a reference or a counter-example. Thank you for your help.

Solution 1:

I don't think this is true. Let $k$ be a field, $Y= \text{Spec } k$, $X= \text{Spec } k[x]_{(x)}$, $f$ the natural morphism, which is clearly flat as $k$ is a field and $\mathcal F = \mathcal O_Y$, so $f^*\mathcal F = \mathcal O_X$. Then $\text{Ass}_Y (\mathcal F) = Y$ but $\text{Ass}_X(f^* \mathcal F)$ is only the point corresponding to the ideal $0$, so it is not equal to $f^{-1}(\text{Ass}_Y (\mathcal F))=X$.

To come up with this example, read the comment I gave. I was thinking if whenever $f :(A, \mathfrak n) \to (B, \mathfrak m)$ is a local, flat homomorphism then $\mathfrak m$ is associated to $B/\mathfrak n B$ so the easiest thing to set is $A$ a field and $B$ a local domain.